The Power of Consecutive Numbers

Algebra Level pending

Suppose 3 consecutive integers are randomly picked and multiplied. Many of you know that the product would just be the cube of the middle number, minus the middle number itself. This can be proved by distributing (x-1), x, and (x+1). But can we expand it…

What is x•(x+1)•(x+2)•(x+3)•(x+4)

Details and Assumptions

In multiple choices, b = (x+2)

b^3 - b^2 b^5 - 5(b^3) + 4b b^6 - 5(b^3) + 3b 120 b^5 - 5(b^3) + 5b b^4 - 5(b^3) + 4b b^5 - 6(b^3) + 4b b^3 - b

Beyond Imagination by Beyond Imagination , 21, USA

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